    Question

# A millivoltmeter of $25$ millivolt range is to be converted into an ammeter of $25$-ampere range. The value (in ohm) of the necessary shunt will be:

A

$1$

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B

$0.01$

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C

$0.05$

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D

$0.001$

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Solution

## The correct option is D $0.001$Step 1. Given data:Range of millivoltmeter = $25$ millivolt = $25×{10}^{-3}$ VoltThe expected range of ammeter $i=25$ ampere Step 2. Calculating the value of shunt resistanceFor making an ammeter, shunt resistance ${R}_{s}$ is connected in parallel to the galvanometer. Here, shunt resistance= ${R}_{s}$, ${V}_{g}$= galvanometer voltage, ${i}_{g}$= full-scale deflection current, Hence ${R}_{S}=\frac{{V}_{g}}{\left(i-{i}_{g}\right)}..........\left(1\right)$, Where subscript g refers to the galvanometer,Now ${i}_{g}=\frac{{V}_{g}}{{R}_{g}}=\frac{25×{10}^{-3}}{{R}_{g}}$Here voltmeter's resistance = ${R}_{g}$Now ${R}_{g}\to \infty ,$ Because in an ideal voltmeter, $R\to \infty$Hence, ${i}_{g}\approx 0$Thus, from $\left(1\right)$,${R}_{s}=\frac{{V}_{g}}{i}=\frac{25×{10}^{-3}}{25}=0.001\Omega$Thus, the correct option is option D.  Suggest Corrections  0      Similar questions  Explore more